Atomic Structure Question 291
Question: If a proton and $ \alpha $ -particle are accelerated through the same potential difference, the ratio of de-Broglie wavelengths $ {\lambda_{p}} $ and $ {\lambda_{\alpha }} $ is
Options:
A) 3
B) $ 2\sqrt{2} $
C) 1
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {\lambda_{p}}=\frac{h}{\sqrt{2eVm_{p}}}; $
$ {\lambda_{H{e^{2+}}}}=\frac{h}{\sqrt{2\times 2eV{m_{H{e^{2+}}}}}}=\frac{h}{\sqrt{2\times 2eV\times 4m_{p}}} $
$ \therefore \frac{{\lambda_{p}}}{{\lambda_{H{e^{2+}}}}}=2\sqrt{2} $
BETA
